Evaluation of an Infinite Series, Revisited

[TMFKAN64]

Since the forum seemed to chew up my post last time, I thought I'd give it another try...

I'm looking to evaluate a series of the form $$\sum_{n=0}^\infty \frac{1}{(n-f)^2}$$ where $$f$$ is a constant between zero and one.

I really have no clue where to start on this. I've seen series such as $$\sum_{n=0}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ solved using a Fourier transform of $$x^2$$, but I can't see how to adapt this.

Any ideas or hints would be appreciated.

 

[mighty2000]

Thank you for your post. Evaluating this series can be a challenging task, but there are a few approaches you can take to solve it. One method is to use the Euler-Maclaurin formula, which can be used to approximate the sum of a series. Another approach is to use the residue theorem from complex analysis, which can help us evaluate the sum using contour integration.

To use the Euler-Maclaurin formula, we first rewrite the series as \sum_{n=0}^\infty \frac{1}{(n-f)^2} = \sum_{n=0}^\infty \frac{1}{n^2} + \sum_{n=0}^\infty \frac{2f}{n^3} + \sum_{n=0}^\infty \frac{f^2}{n^4}. The first sum is the well-known \frac{\pi^2}{6}, while the other two sums can be approximated using the Euler-Maclaurin formula. This will give us an approximate value for the series, which can be refined by increasing the number of terms used in the formula.

Using the residue theorem, we can rewrite the series as \sum_{n=0}^\infty \frac{1}{(n-f)^2} = \sum_{n=0}^\infty \frac{1}{n^2} + \sum_{n=0}^\infty \frac{2f}{n^3} + \sum_{n=0}^\infty \frac{f^2}{n^4} + \sum_{n=0}^\infty \frac{1}{n-f}. The first three sums are the same as before, while the last sum can be evaluated using contour integration. This method may require a bit more mathematical background, but it can give an exact value for the series.

I hope these ideas and hints will help you in evaluating your series. Good luck!